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Question Number 109284 by bobhans last updated on 22/Aug/20

	Answered by abdomsup last updated on 22/Aug/20

	$f (x) = \frac{\sqrt{π - 2 x} - \sqrt{π + 4 x}}{c o s (x - \frac{π}{2})}$ $w e h a v e f (x) = \frac{\sqrt{π - 2 x} - \sqrt{π + 4 x}}{s i n x}$ $\sqrt{π - 2 x} = \sqrt{π} \sqrt{1 - \frac{2 x}{π}} \sim \sqrt{π} (1 - \frac{x}{π})$ $\sqrt{π + 4 x} = \sqrt{π} \sqrt{1 + \frac{4 x}{π}} \sim \sqrt{π} (1 + \frac{2 x}{π})$ $s i n x \sim x \Rightarrow$ $f (x) \sim \frac{\sqrt{π} - \frac{x}{\sqrt{π}} - \sqrt{π} - \frac{2 x}{\sqrt{π}}}{x}$ $= - \frac{3}{\sqrt{π}} \Rightarrow {l i m}_{x \to 0} f (x) = \frac{- 3}{\sqrt{π}}$
	Answered by abdomsup last updated on 22/Aug/20

	$f (x) = \sqrt{8 x^{3} - 2 x^{2}} + 2 x \Rightarrow$ $l i m i t n o t d e f i n e d b e c a u s e$ $8 x^{3} - 2 x^{2} \to - \infty (x \to - \infty)$
	Answered by abdomsup last updated on 22/Aug/20

	$f (x) = {(\frac{x - 2}{x + 2})}^{3 x - 1} \Rightarrow$ $f (x) = e^{(3 x - 1) l n (\frac{x - 2}{x + 2})}$ $= e^{(3 x - 1) l n (1 - \frac{4}{x + 2})} i f x i s r e a l$ $l i m i t n o t d e f i n e d x = 0 \Rightarrow 1 - \frac{4}{x + 2} < 0$ $!$ $i f x c o m p l e x \Rightarrow f (x) = e^{(3 x - 1) (i π) l n (\frac{4}{x + 2} - 1)}$ $\Rightarrow {l i m}_{x \to 0} f (x) = e^{- i π l n (1)} = e^{0} = 1$
	Commented by abdomsup last updated on 22/Aug/20

	$s o r r y f (x) = e^{(3 x - 1) (i π + l n (\frac{4}{x + 2} - 1)}$ $f (x) \to e^{- (i π + l n (1))} = - 1 (x \to 0)$
	Answered by mathmax by abdo last updated on 22/Aug/20

	$f (x) = \frac{{({4 x}^{2} - 2 x)}^{\frac{1}{8}}}{{(8 x + 10)}^{\frac{1}{4}}} \Rightarrow f (x) = \frac{{({4 x}^{2})}^{\frac{1}{8}} {(1 - \frac{1}{2 x})}^{\frac{1}{8}}}{{(8 x)}^{\frac{1}{4}} {(1 + \frac{5}{4 x})}^{\frac{1}{4}}} \sim \frac{(^{8} \sqrt{4})}{(^{4} \sqrt{8})} \times \frac{1 - \frac{1}{16 x}}{1 + \frac{5}{16 x}} (x \to \infty)$ $\Rightarrow \lim_{x \to \infty} f (x) = \frac{(^{8} \sqrt{4})}{(^{4} \sqrt{8})}$
	Answered by mathmax by abdo last updated on 22/Aug/20

	$\lim_{x \to 2^{+}} \frac{∣ x - 2 ∣ \tan (x - 2)}{x^{2} - 4 x + 4} = \lim_{x \to 2^{+}} \frac{∣ x - 2 ∣ \tan (x - 2)}{∣ x - 2 ∣^{2}}$ $= \lim_{x \to 2^{+}} \frac{\tan (x - 2)}{∣ x - 2 ∣} = \lim_{x \to 2^{+}} \frac{x - 2}{x - 2} = 1 (\tan u \sim u atv (o))$ $also \lim_{x \to 2^{-}} \frac{∣ x - 2 ∣ \tan (x - 2)}{x^{2} - 4 x + 4} = \lim_{x \to 2^{-}} \frac{\tan (x - 2)}{2 - x} = - 1$
	Answered by john santu last updated on 22/Aug/20
	$((⊸js⊸)/(⋎⋎⋎⋎)) lim_(x→∞) 5x^2 cot ((8/x)) set (1/x) = j { ((x=(1/j))),((j→0)) :} lim_(j→0) (5/j^2 ) cot (8j) = lim_(j→0) (5/(j^2 tan (8j)))=∞$
	$\frac{⊸ j s ⊸}{⋎ ⋎ ⋎ ⋎}$ $\underset{x \to \infty}{\lim 5} x^{2} \cot (\frac{8}{x})$ $s e t \frac{1}{x} = j {\begin{cases} x = \frac{1}{j} \\ j \to 0 \end{cases}$ $\lim_{j \to 0} \frac{5}{j^{2}} \cot (8 j) = \lim_{j \to 0} \frac{5}{j^{2} \tan (8 j)} = \infty$
	Answered by john santu last updated on 22/Aug/20

	$\Leftrightarrow \frac{⋎ ∙ J S ∙ ⋎}{- - - -}$ $\lim_{x \to 0} {(\frac{x - 2}{x + 2})}^{5 x - 1} = {(- 1)}^{- 1} = \frac{1}{(- 1)} = - 1$
	Commented by mathmax by abdo last updated on 22/Aug/20

	$sir js what s domain of definition of x \to {(\frac{x - 2}{x + 2})}^{5 x - 1} . . . ?$
	Commented by john santu last updated on 22/Aug/20

	$x \neq - 2$
	Commented by mathmax by abdo last updated on 22/Aug/20
	$S =Σ_(n=0) ^∞ ∫_n ^(n+1) (((−1)^([x]) )/(2+cos(n[x]))) dx =Σ_(n=0) ^∞ ∫_n ^(n+1) (((−1)^n )/(2+cos(nx)))dx =Σ_(n=0) ^∞ (−1)^n ∫_n ^(n+1) (dx/(2+cos(nx))) =Σ_(n=0) ^∞ (−1)^n u_n u_n =_(nx=t) ∫_n^2 ^(n^2 +n) (dt/(n(2+cos(t)))) =_(tan((t/2))=u) (1/n) ∫_(tan((n^2 /2))) ^(tan(((n^2 +n)/2))) ((2du)/((1+u^2 )(2+((1−u^2 )/(1+u^2 ))))) =(2/n) ∫_(tan((n^2 /2))) ^(tan(((n^2 +n)/2))) (du/(2+2u^2 +1−u^2 )) =(2/n)∫_(tan((n^2 /2))) ^(tan(((n^2 +n)/2))) (du/(u^2 +3)) =_(u=(√3)z) (2/n) ∫_((1/(√3))tan((n^2 /2))) ^((1/(√3))tan(((n^2 +n)/2))) (((√3)dz)/(3(1+z^2 ))) =(2/(n(√3))){arctan((1/(√3))tan(((n^2 +n)/2))) −arctan((1/(√3))tan((n^2 /2))} ⇒ S =(2/(√3))Σ_(n=0) ^∞ (−1)^n {arctan((1/(√3))tan(((n^2 +n)/2))−arctan((1/(√3))tan((n^2 /2))} rest to study convergence of this serie .....$
	$S = \sum_{n = 0}^{\infty} \int_{n}^{n + 1} \frac{{(- 1)}^{[x]}}{2 + \cos (n [x])} dx = \sum_{n = 0}^{\infty} \int_{n}^{n + 1} \frac{{(- 1)}^{n}}{2 + \cos (nx)} dx$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{n}^{n + 1} \frac{dx}{2 + \cos (nx)} = \sum_{n = 0}^{\infty} {(- 1)}^{n} u_{n}$ $u_{n} =_{nx = t} \int_{n^{2}}^{n^{2} + n} \frac{dt}{n (2 + \cos (t))} =_{\tan (\frac{t}{2}) = u} \frac{1}{n} \int_{\tan (\frac{n^{2}}{2})}^{\tan (\frac{n^{2} + n}{2})} \frac{2 du}{(1 + u^{2}) (2 + \frac{1 - u^{2}}{1 + u^{2}})}$ $= \frac{2}{n} \int_{\tan (\frac{n^{2}}{2})}^{\tan (\frac{n^{2} + n}{2})} \frac{du}{2 + {2 u}^{2} + 1 - u^{2}} = \frac{2}{n} \int_{\tan (\frac{n^{2}}{2})}^{\tan (\frac{n^{2} + n}{2})} \frac{du}{u^{2} + 3}$ $=_{u = \sqrt{3} z} \frac{2}{n} \int_{\frac{1}{\sqrt{3}} \tan (\frac{n^{2}}{2})}^{\frac{1}{\sqrt{3}} \tan (\frac{n^{2} + n}{2})} \frac{\sqrt{3} dz}{3 (1 + z^{2})} = \frac{2}{n \sqrt{3}} {\arctan (\frac{1}{\sqrt{3}} \tan (\frac{n^{2} + n}{2}))$ $- \arctan (\frac{1}{\sqrt{3}} \tan (\frac{n^{2}}{2})} \Rightarrow$ $S = \frac{2}{\sqrt{3}} \sum_{n = 0}^{\infty} {(- 1)}^{n} {\arctan (\frac{1}{\sqrt{3}} \tan (\frac{n^{2} + n}{2}) - \arctan (\frac{1}{\sqrt{3}} \tan (\frac{n^{2}}{2})}$ $rest to study convergence of this serie . . . . .$
	Commented by mathmax by abdo last updated on 22/Aug/20

	$no sir we must have \frac{x - 2}{x + 2} > 0 (its a base)$
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